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Thread: Dependent Events

  1. #1
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    Dependent Events

    A class of students consists of 18 female and 12 male. What would be the probability of selecting 2 students from this class without replacement and selecting one male and one female?

    I know the answer is 0.496 (according to the textbook) but I'm not sure how to obtain this!
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  2. #2
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    Re: Dependent Events

    $$\dfrac{ \dbinom{18}{1}\dbinom{12}{1} }{ \dbinom{30}{2} } = \dfrac{18\cdot 12}{15\cdot 29} \approx 0.496$$

    To get this, first determine the sample space. You have thirty people, and you are randomly choosing two. There are $\dbinom{30}{2} = \dfrac{30\cdot 29}{2\cdot 1} = 435$ total possible outcomes.

    Next, determine the number of outcomes where you wind up with one male and one female:

    Choose one male and choose one female. There are 18 males to choose from and 12 females to choose from. So, the total number of outcomes where you wind up with one male and one female is $18\cdot 12 = 216$.

    Then, to find the probability of that occurring, you divide:

    $\dfrac{216}{435} \approx 0.496$.
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  3. #3
    MHF Contributor

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    Re: Dependent Events

    There are 18 female and 12 male students so 18+ 12= 30 students. The probability that the first student chosen was male is 12/30= 2/5. There are then 18 female and 11 male students. The probability that the second student chosen was female is 18/29. The probability that the students chosen were male and female in that order is (2/5)(18/29)= 36/145.

    The probability that the first student chosen was female is 18/30= 3/5. There are then 17 female and 12 male students. The probability that the second student chosen was male is 12/29. The probability that the students chosen were female and male in that order is (3/5)(12/29)= 36/145 also.

    The probability that the two students chose are male and female in either order is 2(36/145)= 72/145 which, to three decimal places, is 0.496 as before.
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